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(A) Let $A = \begin{bmatrix} 3 & 1 \ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & -1 \ 2 & 3 \end{bmatrix}$. We have $AX = B$,which implies $X

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$\begin{bmatrix} -3 & 4 \ 14 & -13 \end{bmatrix}$

Vedclass · Answer

(A) Let $A = \begin{bmatrix} 3 & 1 \ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & -1 \ 2 & 3 \end{bmatrix}$. We have $AX = B$,which implies $X = A^{-1}B$.First,find the determinant of $A$: $|A| = (3)(1) - (1)(4) = 3 - 4 = -1$.Next,find the adjoint of $A$: $	ext{adj}(A) = \begin{bmatrix} 1 & -1 \ -4 & 3 \end{bmatrix}$.Then,$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-1} \begin{bmatrix} 1 & -1 \ -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \ 4 & -3 \end{bmatrix}$.Now,calculate $X = A^{-1}B = \begin{bmatrix} -1 & 1 \ 4 & -3 \end{bmatrix} \begin{bmatrix} 5 & -1 \ 2 & 3 \end{bmatrix}$.$X = \begin{bmatrix} (-1)(5) + (1)(2) & (-1)(-1) + (1)(3) \ (4)(5) + (-3)(2) & (4)(-1) + (-3)(3) \end{bmatrix} = \begin{bmatrix} -5+2 & 1+3 \ 20-6 & -4-9 \end{bmatrix} = \begin{bmatrix} -3 & 4 \ 14 & -13 \end{bmatrix}$.

If $\begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix} X = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$,then $X =$

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